CODEWITHSUNDEEP
complexity-theorybig-o
Jason
Jason

Reputation: 1613

why log(n)! isn't zero?

I am going to prove that log(n)! is growing faster than log(n!)

I have wondered that why the log(n)! isn't zero for n∈N.

Because I think that log(1) is zero so all following numbers after multiplying the result will become zero.

Thanks in advance.

Upvotes: 0

Views: 128

Answers (2)

Konstantin Dinev
Konstantin Dinev

Reputation: 34905

log(n!) is not zero because:

log(m * n) == log(m) + log(n)
log(2!) = log(2 * 1) = log(2) + log(1) = log(2) + 0

In terms of log(n)! you can ignore the log(1) part. Take for example a binary search tree with one node, its root. The search for a value with a binary search tree is log(n) where n is the number of nodes. So with only a root the search should be zero according to the computation you do, however it is still one, because you have to check the root value. In this sense you can assume log(1) == 1 to avoid confusion.

Upvotes: 1

SomeWittyUsername
SomeWittyUsername

Reputation: 18368

You shouldn't calculate the expression for n that equals to some constant. In order to prove that function f grows faster than function g, you need to show that for n that goes to infinity the value of f will be larger than the value of g.

Upvotes: 1

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